Find points at a given distance on a line of given slope

Difficulty Level : Easy
Last Updated : 29 Oct, 2020

Given the co-ordinates of a 2-dimensional point p(x₀, y₀). Find the points at a distance L away from it, such that the line formed by joining these points has a slope of M.

Examples:

Input : p = (2, 1)
        L = sqrt(2)
        M = 1
Output :3, 2
        1, 0
Explanation:
The two points are sqrt(2) distance away 
from the source and have the required slope
m = 1.

Input : p = (1, 0)
        L = 5
        M = 0
Output : 6, 0
        -4, 0

We need to find two points that are L distance from given point, on a line with slope M.
The idea has been introduced in below post.
Find Corners of Rectangle using mid points

Based on the input slope, the problem can be classified into 3 categories.

If slope is zero, we just need to adjust the x coordinate of the source point
If slope is infinite, the we need to adjust the y coordinate
For other values of slope, we can use the following equations to find the points

$Given \ that \ the \ point (x, y) \ is \ at \ distance \ I \ away \ from \ (x_0, y_0) \newline \newline (y-y_0)^{2} + (x-x_0)^{2}= l^{2} \newline \newline Also \ as \ the \ line \ that \ passes \ through \ (x, y) \ and \ (x0, y0) \ satisfies \newline \newline \frac{y-y_0}{x-x_0}= m \newline \newline Rearranging \ we \ get \newline y=y_0+m*(x-x_0) \newline \newline Putting \ the \ values \ in \ first \ equation \newline \newline m^2.(x-x_0)^2+(x-x_0)^2=l^2 \newline \newline Hence, \ we \ have \newline \newline x=x_0\pm l.\sqrt{\frac{1}{1+m^2}} \newline \newline y=y_0 \pm m.l.\sqrt{\frac{1}{1+m^2}}$

Now using the above formula we can find the required points.

C++

// C++ program to find the points on a line of
// slope M at distance L
#include <bits/stdc++.h>
using namespace std;
 
// structure to represent a co-ordinate
// point
struct Point {
 
    float x, y;
    Point()
    {
        x = y = 0;
    }
    Point(float a, float b)
    {
        x = a, y = b;
    }
};
 
// Function to print pair of points at
// distance 'l' and having a slope 'm'
// from the source
void printPoints(Point source, float l,
                                 int m)
{
    // m is the slope of line, and the
    // required Point lies distance l
    // away from the source Point
    Point a, b;
 
    // slope is 0
    if (m == 0) {
        a.x = source.x + l;
        a.y = source.y;
 
        b.x = source.x - l;
        b.y = source.y;
    }
 
    // if slope is infinte
    else if (m == std::numeric_limits<float>
                                 ::max()) {
        a.x = source.x;
        a.y = source.y + l;
 
        b.x = source.x;
        b.y = source.y - l;
    }
    else {
        float dx = (l / sqrt(1 + (m * m)));
        float dy = m * dx;
        a.x = source.x + dx;
        a.y = source.y + dy;
        b.x = source.x - dx;
        b.y = source.y - dy;
    }
 
    // print the first Point
    cout << a.x << ", " << a.y << endl;
 
    // print the second Point
    cout << b.x << ", " << b.y << endl;
}
 
// driver function
int main()
{
    Point p(2, 1), q(1, 0);
    printPoints(p, sqrt(2), 1);
    cout << endl;
    printPoints(q, 5, 0);
    return 0;
}

Java

// Java program to find the points on 
// a line of slope M at distance L
class GFG{
 
// Class to represent a co-ordinate
// point
static class Point
{
    float x, y;
    Point()
    {
        x = y = 0;
    }
    Point(float a, float b)
    {
        x = a;
        y = b;
    }
};
 
// Function to print pair of points at
// distance 'l' and having a slope 'm'
// from the source
static void printPoints(Point source,
                        float l, int m)
{
     
    // m is the slope of line, and the
    // required Point lies distance l
    // away from the source Point
    Point a = new Point();
    Point b = new Point();
     
    // Slope is 0
    if (m == 0)
    {
        a.x = source.x + l;
        a.y = source.y;
 
        b.x = source.x - l;
        b.y = source.y;
    }
 
    // If slope is infinte
    else if (Double.isInfinite(m))
    {
        a.x = source.x;
        a.y = source.y + l;
 
        b.x = source.x;
        b.y = source.y - l;
    }
    else
    {
        float dx = (float)(l / Math.sqrt(1 + (m * m)));
        float dy = m * dx;
        a.x = source.x + dx;
        a.y = source.y + dy;
        b.x = source.x - dx;
        b.y = source.y - dy;
    }
 
    // Print the first Point
    System.out.println(a.x + ", " + a.y);
 
    // Print the second Point
    System.out.println(b.x + ", " + b.y);
}
 
// Driver code
public static void main(String[] args)
{
    Point p = new Point(2, 1),
          q = new Point(1, 0);
    printPoints(p, (float)Math.sqrt(2), 1);
     
    System.out.println();
     
    printPoints(q, 5, 0);
}
}
 
// This code is contributed by Rajnis09

C#

// C# program to find the points on 
// a line of slope M at distance L
using System;
 
class GFG{
 
// Class to represent a co-ordinate
// point
public class Point
{
    public float x, y;
     
    public Point()
    {
        x = y = 0;
    }
     
    public Point(float a, float b)
    {
        x = a;
        y = b;
    }
};
 
// Function to print pair of points at
// distance 'l' and having a slope 'm'
// from the source
static void printPoints(Point source,
                        float l, int m)
{
     
    // m is the slope of line, and the
    // required Point lies distance l
    // away from the source Point
    Point a = new Point();
    Point b = new Point();
     
    // Slope is 0
    if (m == 0)
    {
        a.x = source.x + l;
        a.y = source.y;
 
        b.x = source.x - l;
        b.y = source.y;
    }
 
    // If slope is infinte
    else if (Double.IsInfinity(m))
    {
        a.x = source.x;
        a.y = source.y + l;
 
        b.x = source.x;
        b.y = source.y - l;
    }
    else
    {
        float dx = (float)(l / Math.Sqrt(
                           1 + (m * m)));
        float dy = m * dx;
        a.x = source.x + dx;
        a.y = source.y + dy;
        b.x = source.x - dx;
        b.y = source.y - dy;
    }
 
    // Print the first Point
    Console.WriteLine(a.x + ", " + a.y);
 
    // Print the second Point
    Console.WriteLine(b.x + ", " + b.y);
}
 
// Driver code
public static void Main(String[] args)
{
    Point p = new Point(2, 1),
          q = new Point(1, 0);
           
    printPoints(p, (float)Math.Sqrt(2), 1);
     
    Console.WriteLine();
     
    printPoints(q, 5, 0);
}
}
 
// This code is contributed by Amit Katiyar

Output:

3, 2
1, 0

6, 0
-4, 0

This article is contributed by Ashutosh Kumar 😀 If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

My Personal Notes

Related Articles

C++

Java

C#